Minimizing Costs (Word Problem)

• Dec 8th 2006, 01:57 PM
tivo1980
Minimizing Costs (Word Problem)
This is a word problem that deals with (minimizing costs):

A power line is to be constructed from a power station at point A to an island at point C, which is 1 mi directly out in the water from a point B on the shore. Point B is 4 mi downshore from the power station at A. It costs $5000 per mile to lay the power line under water and$3000 per mile to lay the line under ground. At what point S downshore from A should the line come to the shore in order to minimize cost? Note that S could very well be B or A (Hint: The length of CS is sqrt(1 + x^2)

• Dec 8th 2006, 03:11 PM
TriKri
Quote:

Originally Posted by tivo1980
This is a word problem that deals with (minimizing costs):

A power line is to be constructed from a power station at point A to an island at point C, which is 1 mi directly out in the water from a point B on the shore. Point B is 4 mi downshore from the power station at A. It costs $5000 per mile to lay the power line under water and$3000 per mile to lay the line under ground. At what point S downshore from A should the line come to the shore in order to minimize cost? Note that S could very well be B or A (Hint: The length of CS is sqrt(1 + x^2)

I guess that $ABC$ forms a right-angle triangle and that $S$ can be chosen anywhere along the line $AB$.

Let's say $x$ is the distance in mi upshores from $B$ to $C$.

The total cost $t\ =\ \sqrt{x^2 +1}\cdot \ 5000 + (4-x)\cdot \ 3000$.

$t\ =\ (x^2 +1)^{0.5}\cdot \ 5000 + (4-x)\cdot \ 3000$.

Minimize $t$:

$\frac{\partial t}{\partial x}\ =\ 0.5\cdot(x^2+1)^{-0.5}\cdot 2\cdot x\cdot \ 5000 - \ 3000\ =\ 0$

$\frac{2 x}{2\sqrt{x^2+1}}\cdot \ 5000\ =\ \ 3000$

$2 x\cdot 5\ =\ 2\sqrt{x^2+1}\cdot 3\ (\Rightarrow x \geq 0)$

$50 x^2\ =\ 36(x^2+1)$

$14 x^2\ =\ 36$

$x^2\ =\ \frac{18}{7}$

$x\ =\ 3\sqrt{\frac{2}{7}}$

Answer: $4-3\sqrt{\frac{2}{7}}$
• Dec 8th 2006, 03:14 PM
tivo1980
Thank you Trikri.
• Dec 8th 2006, 03:23 PM
TriKri
I must be really bored.
• Dec 8th 2006, 03:54 PM
tivo1980
I have a question Trikri. What software did you use to present the math problem the way you did?
• Dec 9th 2006, 12:34 AM
earboth
Quote:

Originally Posted by tivo1980
I have a question Trikri. What software did you use to present the math problem the way you did?

Hello, Tivo,

have a look here: http://www.mathhelpforum.com/math-he...rial-latex.pdf

EB
• Dec 9th 2006, 05:18 AM
earboth
Quote:

Originally Posted by TriKri
I guess that $ABC$ forms a right-angle triangle and that $S$ can be chosen anywhere along the line $AB$.

Let's say $x$ is the distance in mi upshores from $B$ to $C$.

The total cost $t\ =\ \sqrt{x^2 +1}\cdot \ 5000 + (4-x)\cdot \ 3000$.

$t\ =\ (x^2 +1)^{0.5}\cdot \ 5000 + (4-x)\cdot \ 3000$.

Minimize $t$:

$\frac{\partial t}{\partial x}\ =\ 0.5\cdot(x^2+1)^{-0.5}\cdot 2\cdot x\cdot \ 5000 - \ 3000\ =\ 0$

$\frac{2 x}{2\sqrt{x^2+1}}\cdot \ 5000\ =\ \ 3000$

$2 x\cdot 5\ =\ 2\sqrt{x^2+1}\cdot 3\ (\Rightarrow x \geq 0)$

$50 x^2\ =\ 36(x^2+1)$

$14 x^2\ =\ 36$

$x^2\ =\ \frac{18}{7}$

$x\ =\ 3\sqrt{\frac{2}{7}}$

Answer: $4-3\sqrt{\frac{2}{7}}$

Hello, Trikri,

I don't agree with your result:

$2 x\cdot 5\ =\ 2\sqrt{x^2+1}\cdot 3\ (\Rightarrow x \geq 0)$

$100 x^2\ =\ 36(x^2+1)$

$x^2\ =\ \frac{36}{64}$

$x\ =\frac{6}{8}$. The negative result isn't very realistic with this problem.

Answer: $4-\frac{3}{4}=3.25\text{ miles}$ from A.

EB
• Dec 9th 2006, 07:26 AM
TriKri
Quote:

Originally Posted by earboth
Hello, Trikri,

I don't agree with your result:

$2 x\cdot 5\ =\ 2\sqrt{x^2+1}\cdot 3\ (\Rightarrow x \geq 0)$

$100 x^2\ =\ 36(x^2+1)$

Oh, I was a little tired last night ... I must go to bed earlier!

Quote:

$x\ =\frac{6}{8}$. The negative result isn't very realistic with this problem.
That's why I showed that $x \geq 0$
• Dec 9th 2006, 09:15 AM
tivo1980
Thank you earboth for the document and the solution.